Hi everyone,

I am trying to loop through ALL the combinations and count the number
of occurances of the last digit for each of the combinations.
There are 10 categories of last digit :-

111111 2887500
211110 6930000
221100 2772000
222000 105600
311100 924000
321000 316800
330000 3960
411000 39600
420000 3960
510000 396
Total = 13983816

111111 means there are 2,887,500 combinations with all last digits
different.
321000 means there are 316,800 combinations where 3 of the last digits
are the same, 2 of the last digits are the same (but a different last
digit to the 3) and 1 last digit (but a different last digit to the 3
or 2).

Here is what I have so far :-

Option Explicit
Dim A As Long, B As Long, C As Long, D As Long, E As Long, F As Long

Sub LastDigit()
Dim i As Integer
Dim LastDigit As Integer
Dim DigitCounts(0 To 9) As Integer ' This will hold counters for each
digit 0-9
Dim nDupl As Integer
Const minVal As Integer = 1 ' The minimum value in ANY
combination
Const maxVal As Integer = 49 ' The maximum value in ANY
combination

Application.ScreenUpdating = False

For i = 0 To 9
DigitCounts(i) = 0
Next i

For A = minVal To maxVal - 5
For B = A + 1 To maxVal - 4
For C = B + 1 To maxVal - 3
For D = C + 1 To maxVal - 2
For E = D + 1 To maxVal - 1
For F = E + 1 To maxVal

For i = 0 To 4
LastDigit = i - 10 * Int(i) / 10
DigitCounts(LastDigit) = DigitCounts(LastDigit) + 1
Next i

nDupl = 0

For i = 0 To 9
If DigitCounts(i) 1 Then nDupl = nDupl +
DigitCounts(i)
Next i

Next F
Next E
Next D
Next C
Next B
Next A

ActiveCell.Offset(0, 5).Value = nDupl
ActiveCell.Offset(1, 0).Select

Application.ScreenUpdating = True
End Sub

Any help will be greatly appreciated.
Thanks in Advance.
All the Best.
Paul

I don't follow this at all

"111111 means there are 2,887,500 combinations with all last digits
different."

Why ?

What's a combination ?
What's a category ?

321000 means there are 316,800 combinations where 3 of the last digits
are the same, 2 of the last digits are the same (but a different last
digit to the 3) and 1 last digit (but a different last digit to the 3
or 2).

I defy any one to make sense of that !

Why is 6930000 more than 2887500, seems out of sequence ?

In passing, looks like you'll be dealing with numbers over 32k. If
potentially so you should declare them 'As Long' to avoid overflows (As
Integer in VB/VBA is virtually redundant in 32bit systems). Also FWIW, no
need to disable screen updating for the sake of populating just two cells.

Regards,
Peter T



"Paul Black" wrote in message
ups.com...
Hi everyone,

I am trying to loop through ALL the combinations and count the number
of occurances of the last digit for each of the combinations.
There are 10 categories of last digit :-

111111 2887500
211110 6930000
221100 2772000
222000 105600
311100 924000
321000 316800
330000 3960
411000 39600
420000 3960
510000 396
Total = 13983816

111111 means there are 2,887,500 combinations with all last digits
different.
321000 means there are 316,800 combinations where 3 of the last digits
are the same, 2 of the last digits are the same (but a different last
digit to the 3) and 1 last digit (but a different last digit to the 3
or 2).

Here is what I have so far :-

Option Explicit
Dim A As Long, B As Long, C As Long, D As Long, E As Long, F As Long

Sub LastDigit()
Dim i As Integer
Dim LastDigit As Integer
Dim DigitCounts(0 To 9) As Integer ' This will hold counters for each
digit 0-9
Dim nDupl As Integer
Const minVal As Integer = 1 ' The minimum value in ANY
combination
Const maxVal As Integer = 49 ' The maximum value in ANY
combination

Application.ScreenUpdating = False

For i = 0 To 9
DigitCounts(i) = 0
Next i

For A = minVal To maxVal - 5
For B = A + 1 To maxVal - 4
For C = B + 1 To maxVal - 3
For D = C + 1 To maxVal - 2
For E = D + 1 To maxVal - 1
For F = E + 1 To maxVal

For i = 0 To 4
LastDigit = i - 10 * Int(i) / 10
DigitCounts(LastDigit) = DigitCounts(LastDigit) + 1
Next i

nDupl = 0

For i = 0 To 9
If DigitCounts(i) 1 Then nDupl = nDupl +
DigitCounts(i)
Next i

Next F
Next E
Next D
Next C
Next B
Next A

ActiveCell.Offset(0, 5).Value = nDupl
ActiveCell.Offset(1, 0).Select

Application.ScreenUpdating = True
End Sub

Any help will be greatly appreciated.
Thanks in Advance.
All the Best.
Paul

Hi peter,

They are 6 number combinations. So for 6 numbers from 49 numbers there
are 13,983,816 total combinations.
111111 could be numbers 01 02 03 04 05 06, making ALL 6 last digits
different.
321000 could be numbers 01 11 21 30 40 49.

The full list of categories are :-
111111
211110
221100
222000
311100
321000
330000
411000
420000
510000

The program will hopefully calculate the the total combinations for
ALL the categories and list them one under the other.

Thanks in Advance.
All the Best.
Paul

On Oct 5, 10:59 am, "Peter T" <peter_t@discussions wrote:
I don't follow this at all

"111111 means there are 2,887,500 combinations with all last digits
different."

Why ?

What's a combination ?
What's a category ?

321000 means there are 316,800 combinations where 3 of the last digits
are the same, 2 of the last digits are the same (but a different last
digit to the 3) and 1 last digit (but a different last digit to the 3
or 2).

I defy any one to make sense of that !

Why is 6930000 more than 2887500, seems out of sequence ?

In passing, looks like you'll be dealing with numbers over 32k. If
potentially so you should declare them 'As Long' to avoid overflows (As
Integer in VB/VBA is virtually redundant in 32bit systems). Also FWIW, no
need to disable screen updating for the sake of populating just two cells.

Regards,
Peter T

"Paul Black" wrote in message

ups.com...



Hi everyone,

I am trying to loop through ALL the combinations and count the number
of occurances of the last digit for each of the combinations.
There are 10 categories of last digit :-

111111 2887500
211110 6930000
221100 2772000
222000 105600
311100 924000
321000 316800
330000 3960
411000 39600
420000 3960
510000 396
Total = 13983816

111111 means there are 2,887,500 combinations with all last digits
different.
321000 means there are 316,800 combinations where 3 of the last digits
are the same, 2 of the last digits are the same (but a different last
digit to the 3) and 1 last digit (but a different last digit to the 3
or 2).

Here is what I have so far :-

Option Explicit
Dim A As Long, B As Long, C As Long, D As Long, E As Long, F As Long

Sub LastDigit()
Dim i As Integer
Dim LastDigit As Integer
Dim DigitCounts(0 To 9) As Integer ' This will hold counters for each
digit 0-9
Dim nDupl As Integer
Const minVal As Integer = 1 ' The minimum value in ANY
combination
Const maxVal As Integer = 49 ' The maximum value in ANY
combination

Application.ScreenUpdating = False

For i = 0 To 9
DigitCounts(i) = 0
Next i

For A = minVal To maxVal - 5
For B = A + 1 To maxVal - 4
For C = B + 1 To maxVal - 3
For D = C + 1 To maxVal - 2
For E = D + 1 To maxVal - 1
For F = E + 1 To maxVal

For i = 0 To 4
LastDigit = i - 10 * Int(i) / 10
DigitCounts(LastDigit) = DigitCounts(LastDigit) + 1
Next i

nDupl = 0

For i = 0 To 9
If DigitCounts(i) 1 Then nDupl = nDupl +
DigitCounts(i)
Next i

Next F
Next E
Next D
Next C
Next B
Next A

ActiveCell.Offset(0, 5).Value = nDupl
ActiveCell.Offset(1, 0).Select

Application.ScreenUpdating = True
End Sub

Any help will be greatly appreciated.
Thanks in Advance.
All the Best.
Paul- Hide quoted text -

- Show quoted text -

I've narrowed it down to a few different interpretations of what you might
mean. But you win - I give up!

I'm curious now to see if someone else can understand <g

Regards,
Peter T


"Paul Black" wrote in message
oups.com...
Hi peter,

They are 6 number combinations. So for 6 numbers from 49 numbers there
are 13,983,816 total combinations.
111111 could be numbers 01 02 03 04 05 06, making ALL 6 last digits
different.
321000 could be numbers 01 11 21 30 40 49.

The full list of categories are :-
111111
211110
221100
222000
311100
321000
330000
411000
420000
510000

The program will hopefully calculate the the total combinations for
ALL the categories and list them one under the other.

Thanks in Advance.
All the Best.
Paul

On Oct 5, 10:59 am, "Peter T" <peter_t@discussions wrote:
I don't follow this at all

"111111 means there are 2,887,500 combinations with all last digits
different."

Why ?

What's a combination ?
What's a category ?

321000 means there are 316,800 combinations where 3 of the last digits
are the same, 2 of the last digits are the same (but a different last
digit to the 3) and 1 last digit (but a different last digit to the 3
or 2).

I defy any one to make sense of that !

Why is 6930000 more than 2887500, seems out of sequence ?

In passing, looks like you'll be dealing with numbers over 32k. If
potentially so you should declare them 'As Long' to avoid overflows (As
Integer in VB/VBA is virtually redundant in 32bit systems). Also FWIW,
no
need to disable screen updating for the sake of populating just two
cells.

Regards,
Peter T

"Paul Black" wrote in message

ups.com...



Hi everyone,

I am trying to loop through ALL the combinations and count the number
of occurances of the last digit for each of the combinations.
There are 10 categories of last digit :-

111111 2887500
211110 6930000
221100 2772000
222000 105600
311100 924000
321000 316800
330000 3960
411000 39600
420000 3960
510000 396
Total = 13983816

111111 means there are 2,887,500 combinations with all last digits
different.
321000 means there are 316,800 combinations where 3 of the last digits
are the same, 2 of the last digits are the same (but a different last
digit to the 3) and 1 last digit (but a different last digit to the 3
or 2).

Here is what I have so far :-

Option Explicit
Dim A As Long, B As Long, C As Long, D As Long, E As Long, F As Long

Sub LastDigit()
Dim i As Integer
Dim LastDigit As Integer
Dim DigitCounts(0 To 9) As Integer ' This will hold counters for each
digit 0-9
Dim nDupl As Integer
Const minVal As Integer = 1 ' The minimum value in ANY
combination
Const maxVal As Integer = 49 ' The maximum value in ANY
combination

Application.ScreenUpdating = False

For i = 0 To 9
DigitCounts(i) = 0
Next i

For A = minVal To maxVal - 5
For B = A + 1 To maxVal - 4
For C = B + 1 To maxVal - 3
For D = C + 1 To maxVal - 2
For E = D + 1 To maxVal - 1
For F = E + 1 To maxVal

For i = 0 To 4
LastDigit = i - 10 * Int(i) / 10
DigitCounts(LastDigit) = DigitCounts(LastDigit) + 1
Next i

nDupl = 0

For i = 0 To 9
If DigitCounts(i) 1 Then nDupl = nDupl +
DigitCounts(i)
Next i

Next F
Next E
Next D
Next C
Next B
Next A

ActiveCell.Offset(0, 5).Value = nDupl
ActiveCell.Offset(1, 0).Select

Application.ScreenUpdating = True
End Sub

Any help will be greatly appreciated.
Thanks in Advance.
All the Best.
Paul- Hide quoted text -

- Show quoted text -

Definitely strange, but I do see one pattern: All the digits add up to 6.
Subtract one from the right-most 1-digit and add one to the left-most 1-digit
to keep the total at 6, and so on. But the list is missing 600000. Other
than that I'm lost.

"Peter T" wrote:

I've narrowed it down to a few different interpretations of what you might
mean. But you win - I give up!

I'm curious now to see if someone else can understand <g

Regards,
Peter T


"Paul Black" wrote in message
oups.com...
Hi peter,

They are 6 number combinations. So for 6 numbers from 49 numbers there
are 13,983,816 total combinations.
111111 could be numbers 01 02 03 04 05 06, making ALL 6 last digits
different.
321000 could be numbers 01 11 21 30 40 49.

The full list of categories are :-
111111
211110
221100
222000
311100
321000
330000
411000
420000
510000

The program will hopefully calculate the the total combinations for
ALL the categories and list them one under the other.

Thanks in Advance.
All the Best.
Paul

On Oct 5, 10:59 am, "Peter T" <peter_t@discussions wrote:
I don't follow this at all

"111111 means there are 2,887,500 combinations with all last digits
different."

Why ?

What's a combination ?
What's a category ?

321000 means there are 316,800 combinations where 3 of the last digits
are the same, 2 of the last digits are the same (but a different last
digit to the 3) and 1 last digit (but a different last digit to the 3
or 2).

I defy any one to make sense of that !

Why is 6930000 more than 2887500, seems out of sequence ?

In passing, looks like you'll be dealing with numbers over 32k. If
potentially so you should declare them 'As Long' to avoid overflows (As
Integer in VB/VBA is virtually redundant in 32bit systems). Also FWIW,
no
need to disable screen updating for the sake of populating just two
cells.

Regards,
Peter T

"Paul Black" wrote in message

ups.com...



Hi everyone,

I am trying to loop through ALL the combinations and count the number
of occurances of the last digit for each of the combinations.
There are 10 categories of last digit :-

111111 2887500
211110 6930000
221100 2772000
222000 105600
311100 924000
321000 316800
330000 3960
411000 39600
420000 3960
510000 396
Total = 13983816

111111 means there are 2,887,500 combinations with all last digits
different.
321000 means there are 316,800 combinations where 3 of the last digits
are the same, 2 of the last digits are the same (but a different last
digit to the 3) and 1 last digit (but a different last digit to the 3
or 2).

Here is what I have so far :-

Option Explicit
Dim A As Long, B As Long, C As Long, D As Long, E As Long, F As Long

Sub LastDigit()
Dim i As Integer
Dim LastDigit As Integer
Dim DigitCounts(0 To 9) As Integer ' This will hold counters for each
digit 0-9
Dim nDupl As Integer
Const minVal As Integer = 1 ' The minimum value in ANY
combination
Const maxVal As Integer = 49 ' The maximum value in ANY
combination

Application.ScreenUpdating = False

For i = 0 To 9
DigitCounts(i) = 0
Next i

For A = minVal To maxVal - 5
For B = A + 1 To maxVal - 4
For C = B + 1 To maxVal - 3
For D = C + 1 To maxVal - 2
For E = D + 1 To maxVal - 1
For F = E + 1 To maxVal

For i = 0 To 4
LastDigit = i - 10 * Int(i) / 10
DigitCounts(LastDigit) = DigitCounts(LastDigit) + 1
Next i

nDupl = 0

For i = 0 To 9
If DigitCounts(i) 1 Then nDupl = nDupl +
DigitCounts(i)
Next i

Next F
Next E
Next D
Next C
Next B
Next A

ActiveCell.Offset(0, 5).Value = nDupl
ActiveCell.Offset(1, 0).Select

Application.ScreenUpdating = True
End Sub

Any help will be greatly appreciated.
Thanks in Advance.
All the Best.
Paul- Hide quoted text -

- Show quoted text -

Sorry guys,

Charlie you are quite right, I left out the 600000 category, my
appologies.
Basically, there are 13,983,816 combinations of 6 numbers. EACH 6
number combination has a last digit. The program will ideally
calculate each 6 numbers last digit category and keep a count. These
will then be listed.

Thanks in Advance.
All the Best.
Paul

On Oct 5, 6:24 pm, Charlie wrote:
Definitely strange, but I do see one pattern: All the digits add up to 6.
Subtract one from the right-most 1-digit and add one to the left-most 1-digit
to keep the total at 6, and so on. But the list is missing 600000. Other
than that I'm lost.



"Peter T" wrote:
I've narrowed it down to a few different interpretations of what you might
mean. But you win - I give up!

I'm curious now to see if someone else can understand <g

Regards,
Peter T

"Paul Black" wrote in message
roups.com...
Hi peter,

They are 6 number combinations. So for 6 numbers from 49 numbers there
are 13,983,816 total combinations.
111111 could be numbers 01 02 03 04 05 06, making ALL 6 last digits
different.
321000 could be numbers 01 11 21 30 40 49.

The full list of categories are :-
111111
211110
221100
222000
311100
321000
330000
411000
420000
510000

The program will hopefully calculate the the total combinations for
ALL the categories and list them one under the other.

Thanks in Advance.
All the Best.
Paul

On Oct 5, 10:59 am, "Peter T" <peter_t@discussions wrote:
I don't follow this at all

"111111 means there are 2,887,500 combinations with all last digits
different."

Why ?

What's a combination ?
What's a category ?

321000 means there are 316,800 combinations where 3 of the last digits
are the same, 2 of the last digits are the same (but a different last
digit to the 3) and 1 last digit (but a different last digit to the 3
or 2).

I defy any one to make sense of that !

Why is 6930000 more than 2887500, seems out of sequence ?

In passing, looks like you'll be dealing with numbers over 32k. If
potentially so you should declare them 'As Long' to avoid overflows (As
Integer in VB/VBA is virtually redundant in 32bit systems). Also FWIW,
no
need to disable screen updating for the sake of populating just two
cells.

Regards,
Peter T

"Paul Black" wrote in message

oups.com...

Hi everyone,

I am trying to loop through ALL the combinations and count the number
of occurances of the last digit for each of the combinations.
There are 10 categories of last digit :-

111111 2887500
211110 6930000
221100 2772000
222000 105600
311100 924000
321000 316800
330000 3960
411000 39600
420000 3960
510000 396
Total = 13983816

111111 means there are 2,887,500 combinations with all last digits
different.
321000 means there are 316,800 combinations where 3 of the last digits
are the same, 2 of the last digits are the same (but a different last
digit to the 3) and 1 last digit (but a different last digit to the 3
or 2).

Here is what I have so far :-

Option Explicit
Dim A As Long, B As Long, C As Long, D As Long, E As Long, F As Long

Sub LastDigit()
Dim i As Integer
Dim LastDigit As Integer
Dim DigitCounts(0 To 9) As Integer ' This will hold counters for each
digit 0-9
Dim nDupl As Integer
Const minVal As Integer = 1 ' The minimum value in ANY
combination
Const maxVal As Integer = 49 ' The maximum value in ANY
combination

Application.ScreenUpdating = False

For i = 0 To 9
DigitCounts(i) = 0
Next i

For A = minVal To maxVal - 5
For B = A + 1 To maxVal - 4
For C = B + 1 To maxVal - 3
For D = C + 1 To maxVal - 2
For E = D + 1 To maxVal - 1
For F = E + 1 To maxVal

For i = 0 To 4
LastDigit = i - 10 * Int(i) / 10
DigitCounts(LastDigit) = DigitCounts(LastDigit) + 1
Next i

nDupl = 0

For i = 0 To 9
If DigitCounts(i) 1 Then nDupl = nDupl +
DigitCounts(i)
Next i

Next F
Next E
Next D
Next C
Next B
Next A

ActiveCell.Offset(0, 5).Value = nDupl
ActiveCell.Offset(1, 0).Select

Application.ScreenUpdating = True
End Sub

Any help will be greatly appreciated.
Thanks in Advance.
All the Best.
Paul- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -